MHT-CET Maths · Indefinite Integration

Foundations — Antiderivatives, the +C, and Standard Formulae

Integration is differentiation run backwards: given a rate of change, recover the function — plus an unknown constant C that no derivative can pin down.

All Indefinite Integration notes MHT-CET Maths strategy

Why this matters

Before any technique, you need three reflexes: recognise that an indefinite integral is a FAMILY of functions (the +C), recall the standard-formula table cold, and pre-process the integrand with algebra (factor, divide, split) before reaching for a method. 5 PYQs sit directly here — boundary-value problems where a constant must be solved for, and 'reconstruct the function then integrate' shapes — but these reflexes underpin all 121 questions in the chapter.

Concept 1 of 5

Antiderivative and the Constant of Integration

Intuition

Differentiation turns a function into its slope. Integration runs that backwards — given the slope everywhere, find the function. But a constant has zero slope, so ANY vertical shift of a correct answer is also correct. That unknown shift is the +C.

Definition

A function $F$ is an antiderivative of $f$ if $F'(x) = f(x)$ . The indefinite integral $\int f(x)\,dx = F(x) + C$ denotes the whole family of antiderivatives, where $C$ is an arbitrary constant. Because $\dfrac{d}{dx}[F(x) + C] = f(x)$ for every constant $C$ , the constant can never be recovered from $f$ alone — you need an extra condition (a boundary value) to fix it.

Indefinite integral

\int f(x)\,dx = F(x) + C \quad\text{where}\quad F'(x) = f(x)

F(x)any one antiderivative of $f$
Carbitrary constant of integration

Visualization · the +C family of antiderivatives

Every curve is an antiderivative of f(x) = x. They differ only by the constant C — a vertical shift. At x = 1 the red tangents are all parallel (slope = f(1) = 1): same derivative, infinitely many curves. That is why every indefinite integral carries a + C.

Worked example

Verify that

\dfrac{x^2}{2} + 5

and

\dfrac{x^2}{2} - 3

are both antiderivatives of

f(x) = x

Differentiate the first: $\dfrac{d}{dx}\!\left(\dfrac{x^2}{2} + 5\right) = x + 0 = x$ . ✓
Differentiate the second: $\dfrac{d}{dx}\!\left(\dfrac{x^2}{2} - 3\right) = x + 0 = x$ . ✓
Both give back $f(x) = x$ ; they differ only by a constant. So $\int x\,dx = \dfrac{x^2}{2} + C$ .

Answer:Both are valid antiderivatives —

\int x\,dx = \dfrac{x^2}{2} + C

Never drop the +C on an indefinite integral

An indefinite integral with no

+C

is incomplete. MHT-CET options are written so that the 'no constant' version and a wrong-constant version both appear — only

+C

(or

+k

) is correct.

Concept 2 of 5

The Standard-Formula Table

Intuition

Roughly a dozen integrals are the alphabet of the whole chapter. Every technique — substitution, parts, partial fractions — exists only to REDUCE a hard integral to one of these. Know them as reflexes.

Definition

The integrals you must recall instantly:

$\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \neq -1$
$\int \dfrac{1}{x}\,dx = \log|x| + C$
$\int e^x\,dx = e^x + C$ , and $\int a^x\,dx = \dfrac{a^x}{\log a} + C$
$\int \sin x\,dx = -\cos x + C$ , $\int \cos x\,dx = \sin x + C$
$\int \sec^2 x\,dx = \tan x + C$ , $\int \csc^2 x\,dx = -\cot x + C$
$\int \dfrac{dx}{1+x^2} = \tan^{-1}x + C$ , $\int \dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$

Power rule (the most-used row)

\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)

n \neq -1the exclusion that makes $\int x^{-1} = \log|x|$ a separate row

Worked example

Integrate

\displaystyle\int\left(3x^2 + \dfrac{1}{x} + e^x\right)dx

Apply the table term by term (linearity — see the next concept).
$\int 3x^2\,dx = x^3$ ; $\int \dfrac{1}{x}\,dx = \log|x|$ ; $\int e^x\,dx = e^x$ .
Add and attach one constant.

Answer:

x^3 + \log|x| + e^x + C

The power rule excludes $n = -1$

\int x^{-1}\,dx

is NOT

\dfrac{x^0}{0}

— that is undefined. It is the special row

\int \dfrac{1}{x}\,dx = \log|x| + C

. This exclusion is tested directly.

Concept 3 of 5

Linearity and Algebraic Pre-processing

Intuition

Most integrands look hard only because they are written badly. Factor, divide out, split a fraction, or use an identity FIRST — and a 'hard' integral often collapses to the standard table. Always simplify before choosing a method.

Definition

Integration is linear: $\int [a\,f(x) + b\,g(x)]\,dx = a\int f(x)\,dx + b\int g(x)\,dx$ . Pre-processing moves that pay off repeatedly:

Improper fraction (degree of top $\geq$ bottom): do polynomial division first.
Factorable numerator: cancel against the denominator.
Split a single fraction into a sum of simpler ones.

Linearity of integration

\int [a\,f(x) + b\,g(x)]\,dx = a\!\int f(x)\,dx + b\!\int g(x)\,dx

Worked example

Evaluate

\displaystyle\int \dfrac{x^4 - 1}{x^2 + 1}\,dx

Factor the numerator: $x^4 - 1 = (x^2 - 1)(x^2 + 1)$ .
Cancel the denominator: the integrand becomes $x^2 - 1$ .
Integrate term by term: $\int (x^2 - 1)\,dx = \dfrac{x^3}{3} - x + C$ .

Answer:

\dfrac{x^3}{3} - x + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int \dfrac{x^2}{x+1}\,dx

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\int (2x+3)^2\,dx$ — expand first.
2.
$\int \dfrac{x^2-1}{x-1}\,dx$
3.
$\int \dfrac{x+1}{x}\,dx$
4.
$\int \dfrac{x^3}{x^2+1}\,dx$ — set up only (divide).

From the bank · past-year question

Example 3Indefinite IntegrationMODERATE

\int\frac{x^4+x^2+1}{x^2-x+1}\,dx

is equal to

[Q131 · 9th May Shift 1 · 2023]

Divide before you integrate an improper fraction

If the numerator's degree is

\geq

the denominator's, you cannot jump to a log or arctan form. Polynomial-divide first; the remainder term is what carries the log/arctan.

Concept 4 of 5

Finding C from a Boundary Condition

Intuition

When a question gives you

f'(x)

AND one value like

f(1) = 4

, it wants the ONE specific antiderivative, not the family. Integrate to get

F(x) + C

, then plug in the known point to solve for

C

Definition

Given $f'(x)$ and a single value $f(a) = k$ : integrate to obtain $f(x) = F(x) + C$ ; substitute $x = a$ and set equal to $k$ to solve $C = k - F(a)$ . The boundary condition removes the ambiguity of the constant.

Solving for the constant

f(x) = F(x) + C,\qquad C = f(a) - F(a)

f(a) = kthe given boundary value
F(a)antiderivative evaluated at the boundary point

Worked example

f'(x) = 3x^2 + \dfrac{2}{x^3}

and

f(1) = 2

, find

f(x)

Integrate: $\int\!\left(3x^2 + 2x^{-3}\right)dx = x^3 + 2\cdot\dfrac{x^{-2}}{-2} + C = x^3 - \dfrac{1}{x^2} + C$ .
Apply $f(1) = 2$ : $1 - 1 + C = 2$ .
So $C = 2$ .

Answer:

f(x) = x^3 - \dfrac{1}{x^2} + 2

Practice this conceptself-check · 4 quick reps

Try it yourself

\dfrac{d}{dx}f(x) = 4x^3 - \dfrac{3}{x^4}

and

f(2) = 0

, find

f(x)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$f'(x)=2x$ , $f(0)=3$ . Find $f$ .
2.
$f'(x)=\cos x$ , $f(0)=2$ . Find $f$ .
3.
$f'(x)=e^x$ , $f(0)=0$ . Find $f$ .
4.
$f'(x)=\dfrac1x$ , $f(1)=5$ . Find $f$ for $x>0$ .

From the bank · past-year question

Example 4Indefinite IntegrationMODERATE

f'(x) = x - \dfrac{5}{x^5}

and

f(1) = 4

, then

f(x)

[Q109 · 9th May Shift 2 · 2024]

Solve for C only AFTER integrating

The boundary value applies to

f

, not

f'

. Integrate fully first, keep the

C

, then substitute the known point. Substituting into

f'

does nothing.

Drill 1 more on finding c from a boundary condition

Concept 5 of 5

Reconstruct the Function, Then Integrate

Intuition

Some questions hide the integrand: they give a composition like

(f\circ f)(x)

or define

f

implicitly through

f(\text{something}) = \text{expression}

. Build the explicit function FIRST, simplify, and only then integrate.

Definition

Two recurring shapes: (1) composition — compute $(f\circ f)(x) = f(f(x))$ and simplify before integrating; (2) implicit definition — if $f(g(x)) = h(x)$ , substitute $t = g(x)$ , express $x$ in terms of $t$ , and read off $f(t)$ . Then integrate the resulting explicit function using the table.

Composition of a function with itself

(f\circ f)(x) = f\big(f(x)\big)

Worked example

f(x) = \dfrac{1}{1 - x}

and

(f\circ f)(x) = F(x)

, find

\displaystyle\int F(x)\,dx

Compute $f(f(x)) = \dfrac{1}{1 - \frac{1}{1-x}} = \dfrac{1}{\frac{1-x-1}{1-x}} = \dfrac{1-x}{-x} = \dfrac{x-1}{x}$ .
Rewrite for integration: $\dfrac{x-1}{x} = 1 - \dfrac{1}{x}$ .
Integrate: $\int\!\left(1 - \dfrac{1}{x}\right)dx = x - \log|x| + C$ .

Answer:

x - \log|x| + C

Practice this conceptself-check · 4 quick reps

Try it yourself

f\!\left(\dfrac{x-4}{x-2}\right) = 2x + 1

, find

f(x)

explicitly.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$f(x)=2x$ . Find $(f\circ f)(x)$ .
2.
$f(x)=x+1$ . $\int (f\circ f)(x)\,dx$ ?
3.
$f(2x)=4x+1$ . Find $f(x)$ .
4.
$f(x)=\dfrac{1}{x}$ . $(f\circ f)(x)$ ?

From the bank · past-year question

Example 5Indefinite IntegrationMODERATE

f(x)=\frac{x}{x+1},\, x\neq -1

and

(f\circ f)(x) = F(x)

, then

\int F(x)\,dx

[Q144 · 15th May Shift 2 · 2023]

Build f explicitly before integrating

You cannot integrate

f(g(x))

until you know

f

. Substitute to find the explicit rule first; integrating the composition blind is the most common error on these.

Drill 1 more on reconstruct the function, then integrate

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Antiderivative and the Constant of Integration
Indefinite integral
$\int f(x)\,dx = F(x) + C \quad\text{where}\quad F'(x) = f(x)$
The Standard-Formula Table
Power rule (the most-used row)
$\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
Linearity and Algebraic Pre-processing
Linearity of integration
$\int [a\,f(x) + b\,g(x)]\,dx = a\!\int f(x)\,dx + b\!\int g(x)\,dx$
Finding C from a Boundary Condition
Solving for the constant
$f(x) = F(x) + C,\qquad C = f(a) - F(a)$
Reconstruct the Function, Then Integrate
Composition of a function with itself
$(f\circ f)(x) = f\big(f(x)\big)$

Watch out for (5)

Never drop the +C on an indefinite integral
→ Antiderivative and the Constant of Integration
The power rule excludes $n = -1$
→ The Standard-Formula Table
Divide before you integrate an improper fraction
→ Linearity and Algebraic Pre-processing
Solve for C only AFTER integrating
→ Finding C from a Boundary Condition
Build f explicitly before integrating
→ Reconstruct the Function, Then Integrate

Mastery check — 2 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Indefinite IntegrationEASY

\frac{d}{dx}f(x) = 4x^3 - \frac{3}{x^4}

such that

f(2) = 0

, then

f(x)

is equal to

[Q137 · 10th May Shift 2 · 2023]

Example 2Indefinite IntegrationHARD

f\!\left(\frac{x-4}{x-2}\right)=2x+1,\;x\in\mathbb{R}-\{1,-2\}

, then

\int f(x)\,dx

is equal to

[Q124 · 3rd May Shift 2 · 2023]

Drill every past-year question on this subtopic

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